Question

Which of the following species has the highest bond energy?
(A)- $O_2^{2-}$
(B)- $O_2^{+}$
(C)- $O_2^{-}$
(D)- $O_2$

Answer 1

Hint: Bond energy is the energy required to break one mole of a bond to separate the bonded atoms in the gaseous state. It can also be defined as the amount of energy released when one mole of a particular bond is formed between gaseous atoms. Bond energy of a bond is directly proportional to the multiplicity of the bond, i.e. its bond order.

Complete step by step answer:
Bond order tells us the number of covalent bonds present in a molecule. We know that bond energy increases with the bond order. So the molecule having the highest bond order will have the highest bond energy.
 Thus, let us calculate the bond order in the given species.
Bond order is mathematically calculated as half the difference between the number of electrons in the bonding and number of electrons in the antibonding orbitals. If $N_b$ and $N_a$ are the number of bonding and antibonding electrons, respectively, then bond order is given as
     $$\text{Bond order}={\displaystyle \frac{1}{2}}(N_b-N_a)$$

Electronic configuration of oxygen, O (Z=8): $1s^{2}2s^{2}2p_x^{2}2p_y^{1}2p_z^1$.
Electronic configuration of oxygen molecule, $O_2$ :
     $$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^1{\pi }^{\ast }2p_y^1$$

Bond order of peroxide ion $O_2^{2-}$
Oxygen molecule ($O_2$) has 16 electrons. It gains 2 electrons to form $O_2^{2-}$ ion. The two electrons gained will be added into half filled ${\pi }^{\ast }2p_x^1$ and ${\pi }^{\ast }2p_y^1$ such that the electronic configuration of $O_2^{2-}$ is $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^2{\pi }^{\ast }2p_y^2$.

Now, the total number of electrons in bonding molecular orbitals, $N_b$ = 10
Total number of electrons in the antibonding molecular orbitals, $N_a$ = 8
Therefore, the bond order of $O_2^{2-}$ will be
Bond order = ${\displaystyle \frac{1}{2}}(N_a-N_b)$
Bond order = ${\displaystyle \frac{1}{2}}(10-8)=1$

Bond order of $O_2^{+}$ ion.
One electron is lost from ${\pi }^{\ast }2p_x^1$ (or ${\pi }^{\ast }2p_y^1$) molecular orbital of $O_2$ molecule to form $O_2^{+}$ ion. Since, one electron is lost from antibonding molecular orbital, the electronic configuration of $O_2^{+}$ ion becomes
     $$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^1{\pi }^{\ast }2p_y^0$$ .
Total number of electrons in bonding molecular orbitals, $N_b$ = 10
Number of electrons in the antibonding molecular orbitals, $N_a$ = 5
Therefore, bond order = ${\displaystyle \frac{1}{2}}(10-5)=2.5=2{\displaystyle \frac{1}{2}}$

Bond order of $O_2^{-}$ superoxide ion
$O_2$ molecule gains one electron into ${\pi }^{\ast }2p_x^1$ (or ${\pi }^{\ast }2p_y^1$) to form $O_2^{-}$ ion. Now the electronic configuration of $O_2^{-}$ is $\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^2{\pi }^{\ast }2p_y^1$.

Number of bonding electrons, $N_a$ = 10
Number of antibonding electrons, $N_b$ = 7
Therefore, bond order of $O_2^{-}$ = ${\displaystyle \frac{1}{2}}(10-7)=1.5=1{\displaystyle \frac{1}{2}}$

Bond order of oxygen molecule, $O_2$
We know that electronic configuration of $O_2$ molecule is given as
     $$\sigma 1s^2{\sigma }^{\ast }1s^2\sigma 2s^2{\sigma }^{\ast }2s^2\sigma 2p_z^2\pi 2p_x^2\pi 2p_y^2{\pi }^{\ast }2p_x^1{\pi }^{\ast }2p_y^1$$
$O_2$ molecule has 10 electrons in bonding molecular orbitals and 6 electrons in antibonding molecular orbitals, thus its bond is calculated to be
     $${\displaystyle \frac{1}{2}}(N_b-N_a)={\displaystyle \frac{1}{2}}(10-6)=2$$
Therefore, the bond order of the given species follows the order, i.e. $O_2^{+}>O_2>O_2^{-}>O_2^{2-}$.

Since, bond energy increases with the increase of bond order. So, the bond energy of the species also follows the same order as the bond order, i.e.
     $$O_2^{+}>O_2>O_2^{-}>O_2^{2-}$$
Thus, $O_2^{+}$ has the highest bond order, and hence the highest bond energy.

So, the correct answer is “Option B”.

Note: Greater the bond order, higher the bond energy and hence, greater is the stability. Note that the energy of $\sigma 2p_z$ is less than that of $\pi 2p_x$ and $\pi 2p_y$ molecular orbitals and energy of both the $\pi$-orbitals, $\pi 2p_x$ and $\pi 2p_y$ is the same in case of $O_2$, i.e. $\sigma 2p_z<\pi 2p_x=\pi 2p_y$.